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2020-05-27

## 785. Is Graph Bipartite?

### Question:

Given an undirected `graph`, return `true` if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: `graph[i]` is a list of indexes `j` for which the edge between nodes `i` and `j` exists.  Each node is an integer between `0` and `graph.length - 1`.  There are no self edges or parallel edges: `graph[i]` does not contain `i`, and it doesn’t contain any element twice.

#### Example 1:

``````Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
``````

#### Example 2:

``````Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.
``````

Note:

• `graph` will have length in range `[1, 100]`.
• `graph[i]` will contain integers in range `[0, graph.length - 1]`.
• `graph[i]` will not contain `i` or duplicate values.
• The graph is undirected: if any element `j` is in `graph[i]`, then `i` will be in `graph[j]`.

### Solution:

Think as a graph, if we can color all the node with two colors, then the graph is Bipartite. The neighbours node should have a different color than yours. Then using BFS or DFS to check or assign the color to the node next to it.

Using BFS:

``````class Solution {
public boolean isBipartite(int[][] graph) {
if (graph == null || graph.length == 0) return true;

int[] colors = new int[graph.length];
Queue<Integer> que = new LinkedList<>();

for (int i = 0; i < colors.length; i++) {
if (colors[i] != 0) continue;

colors[i] = 1;
que.offer(i);

while (!que.isEmpty()) {
int curr = que.poll();

for (int next : graph[curr]) {
if (colors[curr] == colors[next]) return false;
if (colors[next] == 0) {
colors[next] = -colors[curr];
}
}
}
}
return true;
}
}
``````

Using DFS:

``````class Solution {
public boolean isBipartite(int[][] graph) {
if (graph == null || graph.length == 0) return true;

int[] colors = new int[graph.length];
for (int i = 0; i < colors.length; i++) {
if (colors[i] == 0 && !dfs(i, 1, graph, colors)) return false;
}

return true;
}

private boolean dfs(int i, int targetcolor, int[][] graph, int[] colors){
if (colors[i] != 0) return colors[i] == targetcolor;

colors[i] = targetcolor;

if (graph[i] == null) return true;

for (int next: graph[i]) {
if (!dfs(next, -targetcolor, graph, colors)) return false;
}

return true;
}
}
``````

Time complexity: O(V+E)
Space complexity: O(V)